Parallel Lines in a Cyclic Quadrilateral

What is this about?
A Mathematical Droodle


This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at http://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

img src="PLICQ.gif" width=279 height=245 alt="Parallel Lines in a Cyclic Quadrilateral">


What if applet does not run?

Explanation

|Activities| |Contact| |Front page| |Contents| |Geometry| |Store|

Copyright © 1996-2017 Alexander Bogomolny

The applet attempts to suggest a problem from the 1993-1994 Saint Petersburg (Russia) Regional Mathematical Olympiad for grade 9

Two crossing chords AC and BD are drawn in a given circle. M is a point on AC such that AM = AD. N is a point on BD such that BN = BC. Prove that, provided M and N are distinct, MN||AB.


This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at http://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

img src="PLICQ.gif" width=279 height=245 alt="Parallel Lines in a Cyclic Quadrilateral">


What if applet does not run?

Let G be the intersection of AC and BD. We shall focus on a pair of similar triangle: AGD and BGC. Importantly, two inscribed angles are equal - ∠CAD = ∠CBD - as subtended by the same arc.

Triangles AGD and BGC are similar because ∠GAD = ∠CAD = ∠CBD = ∠CBG, and also ∠AGD = ∠BGC, as vertical. This gives us a proportion:

AG/AD = BG/BC.

Since AD = AM and BC = BN, the proportion could be rewritten as

AG/AM = BG/BN.

Triangles ABG and MNG have two pairs of proportional sides and equal angles in-between. They are therefore similar and have pairwise equal angles. The lines AB and MN are parallel because, say, equal angles GMN and GAB are, depending on the configuration, either corresponding or alternate with respect to the transversal AM.

Note that the requirement that the chords AD and BC cross is redundant. Both the problem and the solution stand even if they do not.

Angles in Circle

  1. Angle Subtended by a Diameter
  2. Inscribed Angles
  3. Inscribed and Central Angles in a Circle
  4. Munching on Inscribed Angles
  5. Sangaku with Angle between a Tangent and a Chord
  6. Secant Angles in a Circle
  7. Secant Angles in a Circle II
  8. Thales' theorem

|Activities| |Contact| |Front page| |Contents| |Geometry| |Store|

Copyright © 1996-2017 Alexander Bogomolny

 62056788

Search by google: